What is the volume of 5.0 grams of CO_2 at STP?

2 Answers
May 18, 2017

The volume of "5.0 g CO"_2" is "2.6 L CO"_2" at STP.

Explanation:

color(blue)("STP"

STP is currently 0^@"C" or "273.15 K", which are equal, though the Kelvin temperature scale is used for gas laws; and pressure is "10"^5color(white)(.)"Pascals (Pa)", but most people use "100 kPa", which is equal to 10^5color(white)(.)"Pa".

You will use the ideal gas law to answer this question. Its formula is:

PV=nRT,

where P is pressure, V is volume, n is moles, R is a gas constant, and T is temperature in Kelvins.

color(blue)("Determine moles"
You may have noticed that the equation requires moles (n), but you have been given the mass of "CO"_2". To determine moles, you multiply the given mass by the inverse of the molar mass of "CO"_2", which is "44.009 g/mol".

5.0color(red)cancel(color(black)("g CO"_2))xx(1"mol CO"_2)/(44.009color(red)cancel(color(black)("g CO"_2)))="0.1136 mol CO"_2"

color(blue)("Organize your data".

Given/Known
P="100 kPa"
n="0.1136 mol"
R="8.3145 L kPa K"^(-1) "mol"^(-1)"
https://en.wikipedia.org/wiki/Gas_constant
T="273.15 K"

Unknown: V

color(blue)("Solve for volume using the ideal gas law."
Rearrange the formula to isolate V. Insert your data into the equation and solve.

V=(nRT)/P

V=(0.1136color(red)cancel(color(black)("mol"))xx8.3145 color(white)(.)"L" color(red)cancel(color(black)("kPa")) color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx273.15color(red)cancel(color(black)("K")))/(100color(red)cancel(color(black)("kPa")))="2.6 L CO"_2" rounded to two significant figures due to "5.0 g"

May 18, 2017

I got 2.55 Liters

Explanation:

1 mole of any gas at STP = 22.4 Liters
5 g CO_2(g) = (5 g)/(44(g/"mole")) = 0.114 "mole" CO_2(g)
Volume of 0.114 mole CO_2(g) = (0.114 mole)(22.4 L/mole) = 2.55 Liters CO_2(g) at STP